\newproblem{lay:5_2_19}{
  % Problem identification
	\begin{large}
	  \hspace{\fill}\newline
    \textbf{Lay, 5.2.19}
	\end{large}
	\\
  \ifthenelse{\boolean{identifyAuthor}}{\textit{Carlos Oscar Sorzano, Aug. 31st, 2013} \\}{}

  % Problem statement
	Let $A$ be an $n\times n$ matrix, and suppose $A$ has $n$ real eigenvalues, $\lambda_1, \lambda_2, ..., \lambda_n$, repeated according to multiplicities, so that
	\begin{center}
		$\det\{A-\lambda I\}=(\lambda_1-\lambda)(\lambda_2-\lambda)...(\lambda_n-\lambda)$
	\end{center}
	Explain why $\det\{A\}$ is the product of the $n$ eigenvalues of $A$. (This result is true for any square matrix when complex eigenvalues are considered.)
}{
   % Solution
	Since the equation above is true for any value of $\lambda$, we simply take $\lambda=0$ to obtain
	\begin{center}
		$\det\{A\}=\lambda_1\lambda_2...\lambda_n$
	\end{center}
}
\useproblem{lay:5_2_19}
\ifthenelse{\boolean{eachProblemInOnePage}}{\newpage}{}
